Prove that Cl(Q)=R in the standard topology on R.
Let $\displaystyle a\in R$ for any any $\displaystyle \epsilon > 0$ we know that $\displaystyle (a-\epsilon,a+\epsilon)$ contains both rational and irrational points. Therefore, all points are boundary points. Thus, $\displaystyle \partial \mathbb{Q} = \mathbb{R}$ which means $\displaystyle \text{cl}( \mathbb{Q}) = \mathbb{Q} \cup \partial \mathbb{Q} = \mathbb{R}$.